Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 246: 80


$$\frac{1}{3} \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \frac{1}{2}$$

Work Step by Step

Since$$ f^{\prime}(x)=\frac{15 * x^{2}}{\sqrt{5 x^{3}+4}} \geq 0$$ then $f(x)$ is increasing on $ [ 0,1] $ and $ \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}}$ has an upper bound at $x=0$ $$\int_{0}^{1} \frac{1}{\sqrt{5(0)+4}}=\int_{0}^{1} \frac{1}{2} d x=1 / 2$$ and a lower bound $$\int_{0}^{1} \frac{1}{\sqrt{5(1)+4}}=\int_{0}^{1} \frac{1}{3} d x=1 / 3$$ Then, by using the comparison theorem \begin{aligned} \int_{0}^{1} \frac{1}{3} d x & \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \int_{0}^{1} \frac{1}{2} d x \\ \frac{1}{3} *(1-0) & \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \frac{1}{2} *(1-0) \\ \frac{1}{3} & \leq \int_{0}^{1} \frac{d x}{\sqrt{5 x^{3}+4}} \leq \frac{1}{2} \end{aligned}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.